Kinematics Problem

[jonnburton]

Hi,

I wondered if anyone could give me some help with this problem that I am unable to solve.

Two cars, A and B are initially at rest side by side. A sets off on a straight track with an acceleration of 2m/s. Five seconds later B starts off on a parallel track to A with acceleration 3.125m/s.


The first part was to calculate the distance travelled by A in 5 seconds, which I managed to do, and found the result is 25m.


The next part is where I am having difficulties.:

Calculate the time taken for B to catch up with A.

This is what I did:

\(\displaystyle s = ut + \frac{1}{2} at^2\)


Car A

\(\displaystyle s = t^2\)


Car B

Given that it started five seconds later than A and A moved 25 metres in that time, I think this is the same as saying B has to move 25 metres more than A. So:

\(\displaystyle s +25 = \frac{1}{2}*3.125t^2\)

\(\displaystyle s = \frac{25}{16} t^2 -25\)



Combining these results : B-A:

\(\displaystyle \frac{25}{16} t^2 -25 -t^2 = 0\)

\(\displaystyle \frac{9}{16} t^2 -25 = 0\)

\(\displaystyle \frac{9}{16}t^2 = 25\)

\(\displaystyle 9t^2 = 400\)

\(\displaystyle t^2 = 44.4\)

\(\displaystyle t =6.7 sec\)


According to the book the answer is 20 seconds, though..!

 

[MarkFL]

Let's start the clock at the moment car B begins to move. At this time, car A has moved 25 m and its velocity is 10 m/s (this is what you neglected to use). So, we want to write:

\(\displaystyle \displaystyle \frac{1}{2}2t^2+10t+25=\frac{1}{2}\frac{25}{8}t^2\)

Arranging in standard quadratic form, we find:

\(\displaystyle \displaystyle \frac{9}{16}t^2-10t-25=0\)

\(\displaystyle 9t^2-160t-400=0\)

\(\displaystyle (t-20)(9t+20)=0\)

Discarding the negative root, we find:

\(\displaystyle t=20\)

 

[Deleted member 4993]

Hi,

I wondered if anyone could give me some help with this problem that I am unable to solve.

Two cars, A and B are initially at rest side by side. A sets off on a straight track with an acceleration of 2m/s. Five seconds later B starts off on a parallel track to A with acceleration 3.125m/s.


The first part was to calculate the distance travelled by A in 5 seconds, which I managed to do, and found the result is 25m.


The next part is where I am having difficulties.:

Calculate the time taken for B to catch up with A.

This is what I did:

\(\displaystyle s = ut + \frac{1}{2} at^2\)


Car A

\(\displaystyle s = t^2\)


Car B

Given that it started five seconds later than A and A moved 25 metres in that time, I think this is the same as saying B has to move 25 metres more than A. So:

\(\displaystyle s +25 = \frac{1}{2}*3.125t^2\)

\(\displaystyle s = \frac{25}{16} t^2 -25\)



Combining these results : B-A:

\(\displaystyle \frac{25}{16} t^2 -25 -t^2 = 0\)

\(\displaystyle \frac{9}{16} t^2 -25 = 0\)

\(\displaystyle \frac{9}{16}t^2 = 25\)

\(\displaystyle 9t^2 = 400\)

\(\displaystyle t^2 = 44.4\)

\(\displaystyle t =6.7 sec\)


According to the book the answer is 20 seconds, though..!

Assume that at time 't' A & B caught up with each other. That means at that time A & B have traveled the same distance.

1/2 * 2 * (t + 5)² = 1/2 * 3.125 * (t)²

t² + 10t + 25 = 1.5625t²

0.5625 * t²- 10t - 25 = 0

Solve for 't' .....

 

[jonnburton]

Thank you for your help MarkFL and Subhotosh. Now I see where my thinking was going wrong.

 

[HallsofIvy]

Hi,

I wondered if anyone could give me some help with this problem that I am unable to solve.

Two cars, A and B are initially at rest side by side. A sets off on a straight track with an acceleration of 2m/s. Five seconds later B starts off on a parallel track to A with acceleration 3.125m/s.


The first part was to calculate the distance travelled by A in 5 seconds, which I managed to do, and found the result is 25m.


The next part is where I am having difficulties.:

Calculate the time taken for B to catch up with A.

This is what I did:

\(\displaystyle s = ut + \frac{1}{2} at^2\)


Car A

\(\displaystyle s = t^2\)


Car B

Given that it started five seconds later than A and A moved 25 metres in that time, I think this is the same as saying B has to move 25 metres more than A. So:

\(\displaystyle s +25 = \frac{1}{2}*3.125t^2\)

\(\displaystyle s = \frac{25}{16} t^2 -25\)

This is your error. You can either take t= 0 at the instant B starts moving and have A start at the same time from 25 m ahead of B: (1/2)(2)t^2+ 25= (1/2)(3.125)t^2, or take t= 0 at the instant A starts moving, so that B starts at t= 5 and moves for time t- 5, and have A and B cover the same distance: (1/2)(2)t^2= (1/2)(3.125)(t- 5)^2.

But don't do both!

Combining these results : B-A:

\(\displaystyle \frac{25}{16} t^2 -25 -t^2 = 0\)

\(\displaystyle \frac{9}{16} t^2 -25 = 0\)

\(\displaystyle \frac{9}{16}t^2 = 25\)

\(\displaystyle 9t^2 = 400\)

\(\displaystyle t^2 = 44.4\)

\(\displaystyle t =6.7 sec\)


According to the book the answer is 20 seconds, though..!